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An amplification root is a root of a polynomial governing a geometric mode $y_n={\xi }^n$ of a numerical recurrence on the Dahlquist test equation. Its modulus measures damping or growth per step. A repeated root produces additional modes such as $n{\xi }^n$; therefore a unit-circle root can be permitted by a bounded-root condition only when it is simple. The principal root approximates $e^z$ near zero, while additional roots can describe parasitic amplification roots.

The stability convention matters. The Cambridge numerical analysis notes define a strict linear stability domain through decay $y_n\to 0$. For a multistep recurrence, robust decay requires every amplification root to lie strictly inside the unit disk. With that convention we can prove the stronger result that this method's strict linear stability domain is empty, and therefore bounded. Mere boundedness of oscillatory sequences gives a different answer, described below.

On the Dahlquist test equation, $g(y)={\lambda }^{2}y$. With $z=h\lambda$, the amplification polynomial of a multistep method isExclude $D=0$, where the intended two-step update cannot be solved. If both polynomial roots have modulus below one, their product requires $|A|<|D|$. Butso necessarily $\mathrm{Re}z<0$. The complex quadratic Schur criterion also requiresIts left-hand side equals $32|\mathrm{Re}z|(|z{|}^2+15)$, while for $\mathrm{Re}z<0$ its right-hand side is only $28|\mathrm{Re}z|(|z{|}^2+15)$. The inequality is impossible. ThusFor comparison, the bounded-root stability set of the symmetric two-derivative formula is unbounded. On $z=iy$, rotate the amplification polynomial into a real reciprocal quadratic. Its two roots are distinct and on the unit circle precisely whenequivalently $y^2<(45-15\sqrt{5})/2$ or $y^2>(45+15\sqrt{5})/2$. In particular, arbitrarily large imaginary parameters give bounded, undamped numerical oscillations. At the two equality thresholds, repeated unit-circle roots produce unbounded solutions. Hence the question's bounded-domain claim holds for the strict decay convention; it does not hold if the phrase is interpreted using the non-growing root condition for a multistep method.

A constant-step linear multistep method has the formwith ${\alpha }_k\ne 0$. The characteristic polynomials of a linear multistep method encode both accuracy and error propagation. The method has order of a numerical method $p$ when insertion of a sufficiently smooth exact solution leaves an unnormalized local truncation error $O(h^{p+1})$. The exponential-symbol order criterion for a multistep method expresses this as $\rho (e^z)-z\sigma (e^z)=O(z^{p+1})$. In particular, first-order consistency of a numerical method requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$.

Accuracy alone is insufficient. For $f=0$, perturbations obey the recurrence $\rho (E)e_n=0$. The root condition for a multistep method requires every root of a polynomial $\rho$ to lie in the closed unit disk, with each unit-circle root simple. This is zero-stability. A root outside the unit disk causes geometric amplification as $n\sim T/h$ increases; a repeated unit-circle root permits polynomial growth in $n$. Either can destroy convergence even for starting perturbations tending to zero.

The Dahlquist equivalence theorem states that, for a fixed-coefficient method and a well-posed ordinary differential equation with a suitably Lipschitz continuous right-hand side, consistency of a numerical method and zero-stability are equivalent to convergence for all convergent starting data. If the starting errors are $O(h^p)$ and the exact solution is sufficiently smooth, the global error is $O(h^p)$ on a fixed time interval. A stability estimate bounds the propagated sum of $O(h^{p+1})$ local defects over $O(1/h)$ steps; the discrete Gronwall inequality controls the dependence of $f$ on the accumulated error. Implicit methods also require the locally consistent, uniquely solvable update branch for sufficiently small $h$.

For example, the explicit Euler method has order one and $\rho (\zeta )=\zeta -1$, satisfying the root condition for a multistep method. The two-step Adams-Bashforth method has order two and $\rho (\zeta )=\zeta (\zeta -1)$, again satisfying it. The trapezoidal rule has order two with one step. The two-step Simpson multistep method, $Y_{n+2}-Y_n=h(f_{n+2}+4f_{n+1}+f_n)/3$, has order four and the simple roots $1,-1$: it is zero-stable, although its parasitic amplification root can spoil long-time decay. These examples distinguish finite-time numerical convergence from absolute stability.

The first Dahlquist barrier gives the maximum possible order of a convergent ordinary real $k$-step method:These bounds assume only first-derivative evaluations, fixed coefficients and zero-stability; the sixth-order multiderivative multistep method in solution is outside that class.

To obtain arbitrarily high convergent order within the ordinary class, increase the number of steps. The Adams-Bashforth method integrates a polynomial interpolation of $k$ past derivative values and has order $k$. The Adams–Moulton method includes the new endpoint and has order $k+1$ for $k$ steps. In both families $\rho (\zeta )={\zeta }^{k-1}(\zeta -1)$, so the root condition for a multistep method holds for every fixed $k$. Starting values can be generated by a Runge-Kutta method of sufficient accuracy. The backward differentiation formula instead differentiates an interpolating polynomial; its order-$k$ members are zero-stable only for $1\le k\le 6$.

Finally, high order and favorable stiff stability are separate design goals. By the Second Dahlquist barrier, an ordinary A-stable linear multistep method has order at most two. One may accept a smaller linear stability domain, use suitable backward differentiation formula members for stiff problems, or switch to high-order implicit Runge-Kutta methods or multiderivative multistep methods. A convergent high-order formula still needs time steps appropriate to its stability properties and the problem's eigenvalues.

The strict decay convention for a linear stability domain demands that every numerical mode tends to zero on the Dahlquist test equation. For a one-step stability function it requires $|R(z)|<1$; for a multistep method it requires every amplification root to have modulus below one. This differs from a boundedness convention that allows simple unit-circle roots, and from mesh-uniform finite-time numerical stability. Cambridge numerical analysis notes use the decay convention. Explicitly specifying the convention is essential for methods with undamped modes.